Question
Answer
$$(-1+i)\cdot(1-i)\cdot\dfrac{1}{2}=\dfrac{2i}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(-1+i)\cdot(1-i)\cdot\frac{1}{2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(-1+i+i-i^2)\cdot\frac{1}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(-i^2+2i-1)\cdot\frac{1}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(1+2i-1)\cdot\frac{1}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2i\cdot\frac{1}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{2i}{2}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{-1+i}\right) $ by each term in $ \left( 1-i\right) $. $$ \left( \color{blue}{-1+i}\right) \cdot \left( 1-i\right) = -1+i+i-i^2 $$ |
| ② | Combine like terms: $$ -1+ \color{blue}{i} + \color{blue}{i} -i^2 = -i^2+ \color{blue}{2i} -1 $$ |
| ③ | $$ -i^2 = -(-1) = 1 $$ |
| ④ | Combine like terms: $$ \, \color{blue}{ \cancel{1}} \,+2i \, \color{blue}{ -\cancel{1}} \, = 2i $$ |
| ⑤ | Multiply $2i$ by $ \dfrac{1}{2} $ to get $ \dfrac{ 2i }{ 2 } $. Step 1: Write $ 2i $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 2i \cdot \frac{1}{2} & \xlongequal{\text{Step 1}} \frac{2i}{\color{red}{1}} \cdot \frac{1}{2} \xlongequal{\text{Step 2}} \frac{ 2i \cdot 1 }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2i }{ 2 } \end{aligned} $$ |
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